Question: $ g(x) = \int_{2}^{x} {\frac{1}{{1 + {t^{\,3}}}}\,dt} \,$ $ g\,^\prime(2)\, =$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{-7}{18}$ (Choice B) B $0$ (Choice C) C $\dfrac{1}{9}$ (Choice D) D $\dfrac{1}{3}$ (Choice E) E None of these
Explanation: The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = \dfrac{1}{{1 + {t^{\,3}}}}$ is continuous on $[2,2]$. Applying the theorem We're given: $ g(x) = \int_{2}^{x} {\frac{1}{{1 + {t^{\,3}}}}\,dt} $ So the theorem tells us: $ g\,^\prime(x) = \frac{1}{{1 + {x^{\,3}}}} $ Evaluating $g'\left(2\right)$ $ g\,^\prime(2) = \frac{1}{{1 + 2^3}}=\dfrac{1}{9}$ The answer: $g'\left(2\right)=\dfrac19$